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Integration cos^2(∏/2cosθ)by doey
Tags: integration 
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#1
Dec312, 09:16 AM

P: 27

i facing a maths problem in integrating ∫ cos^2(∏/2cosθ) with limit from 0 to ∏/2,i was panic and struggled a long period of time in solving this,anyone can help me? pls give me the answer in detail tq !



#2
Dec312, 10:13 AM

P: 295

Use the integral identity [itex]\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(ax)\,dx[/itex].



#3
Dec312, 10:32 AM

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Why would that help, Millenial?
You change the internal cos(theta) to a sin(theta).. 


#4
Dec312, 10:40 AM

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Integration cos^2(∏/2cosθ)
(1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex] or (2) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2\cos \theta}) d\theta[/itex] Also, which methods do you have at your disposal? Contour integration? Differentiation under the integral sign? Just normal calc II techniques? 


#5
Dec312, 11:02 AM

P: 27

,i am asking this pls let me know the steps it takes 


#6
Dec312, 11:21 AM

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#7
Dec312, 11:33 AM

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Just a thought:
We may easily rewrite this equation into the identity: [tex]\int_{0}^{\frac{\pi}{2}}\cos^{2}(\frac{\pi}{2}\cos\theta)d\theta+\int_{ 0}^{\frac{\pi}{2}}\sin^{2}( \frac{\pi}{2}\cos\theta)d\theta=\frac{\pi}{2}[/tex] I feel dreadfully tempted to declare the two integrals to have the same value (the latter being merely a flipped version of the first), but temptation is not proof.. 


#8
Dec312, 11:39 AM

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Anyway, wolfram alpha gives us [itex]\int_0^{\pi/2} \cos^2(\frac{\pi}{2} \cos(x))dx = \frac{\pi}{4}(1+J_0(\pi))[/itex] so I doubt the integral will be solvable with methods like these. 


#9
Dec312, 04:32 PM

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P: 1,391

To evaluate the integral one will have to use the identities
$$\cos t = \frac{1}{2}(e^{it}+e^{it})$$ (or just ##\cos t = \mbox{Re}[\exp(it)]##) and $$e^{iz\cos\theta} = \sum_{n=\infty}^\infty i^n J_n(z)e^{in\theta}.$$ I guess the trig identity $$\cos^2 t = \frac{1}{2}(1+\cos(2t))$$ also helps. 


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